Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 5 - The Integral - 5.4 The Fundamental Theorem of Calculus, Part I - Exercises - Page 258: 23


$$60\sqrt 3 -\frac{8}{3} $$

Work Step by Step

We have $$\int_{1}^{27}(t+1) t^{-1/2} d t=\int_{1}^{27}t^{1/2}+t^{-1/2} d t\\ =\frac{t^{3/2}}{3/2}+\frac{t^{1/2}}{1/2} |_{1}^{27}=\frac{27 \sqrt 3}{3/2}+\frac{3\sqrt 3}{1/2}-\frac{2}{3}-2\\ = 60\sqrt 3 -\frac{8}{3} $$
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